Question

A particle is projected at $${ 60 }^{ }$$ to the horizontal with a kinetic energy KE. The kinetic energy at the highest point is :

A
KE
B
Zero
C
KE4
D
KE2

Solution

The correct option is C $$\dfrac { KE }{ 4 }$$Let initial velocity of the particle be $$v$$.$$\therefore KE=\dfrac { 1 }{ 2 } { mv }^{ 2 }$$at the highest point of its flight    $$v=u\cos { { 60 }^{ } } =\dfrac { u }{ 2 }$$$$\therefore$$ Kinetic energy at this point   $$=\dfrac { 1 }{ 2 } { mv }^{ 2 }=\dfrac { 1 }{ 2 } \dfrac { { mu }^{ 2 } }{ 4 } =\dfrac { KE }{ 4 }$$Physics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More